If AABC is reflected across the y-axis, what are the coordinates of A'?

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 1.

<h2> $y = \frac{2}{x} + 3$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1

, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

<h2> $y = {2}^{x} - 1$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

<h2> $y = {x}^{2} + 7$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 2.

<h3>Hope it is helpful...</h3>

6 0

2 years ago

The diffrence between 2 numbers is 1 and their product Is 132

il63 [147K]

Answer:

x = 12

y = 11

For any clarification, Please contact me in the comment section :)

I Hope It Helps .

Step-by-step explanation:

$Let -the-unkonown- numbers-be-x-and-y \\x-y = 1 (1)\\xy = 132 (2)\\Make-x-subject-of-formula-in-equation1 \\x = 1+y\\Substitute , 1+y for- x -in -equation ,2\\y(1+y) = 132\\y + y^{2} =132\\y^{2} +y-132=0\\Solve- the-quadratic-equation.\\y = 11\\substitute- 11 -for- y- in- equation -1- 0r- 2\\x -y = 1\\x-11 = 1\\x = 1+11\\x = 12$

6 0

3 years ago

Find the area of the shaded regions below. Give your answer as a completely simplified exact value in terms of π (no approximati

pochemuha

!

Answer: 8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2

Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

= 4 so the semicircle area is

(1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

First of all, angle ACB is a right angle ( i.e. 90 degrees).

* This is the Theorem of Thales from elementary Plane Geometry. *

so by Pythagoras

AC^2 + BC^2 = AB^2

But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

Substituting these in Pythagoras gives

AC^2 + 4^2 = 8^2 or

AC^2 = 8^2 - 4^2- = 64 - 16 = 48

Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

(1/2) b * h = (1/2) BC * AC = (1/2) 4 * (4 * sqrt(3)) = 8 sqrt(3)

We conclude the area area of the shaded part is

8 PI - 8 sqrt(3) = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx 1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

8 0

2 years ago